Struct Multidisc Optim 22, 394–406 Springer-Verlag 2001
Optimal plastic design of a bar under combined torsion, bending and shear ˙ W. Egner and M. Zyczkowski
Abstract Optimal plastic design of a bar or beam under torsion with superposed bending and shear is obtained via the boundary perturbation method. Effects of bending and shear are expressed by a small parameter α. The results of optimization of solid bars and of hollow bars are shown. Key words plastic design, combined loadings, solid bars, hollow bars
1 Introduction In the present paper we consider optimal plastic design of a bar under torsion with superposed simultaneous small bending and shear. This problem will be solved by means of the boundary perturbation method (BPM) with a bar of constant circular or annular section assumed as the basic solution (optimal in the case of pure torsion) and a small perturbing parameter describing the effect of bending and shear. The boundary perturbation method is now widely used in various branches of solid and fluid mechanics. BPM was applied to determination of shapes of bodies showing full plastification at the stage of collapse first ˙ by Kordas and Zyczkowski (1970). Full plastification, if possible at all, was then regarded as the first step towards subsequent optimal plastic design. Ample survey of the relevant literature was given in the paper by Egner et al. (1994) and will not be repeated here. We mention ˙ just some more recent related papers by Zyczkowski and Egner (1995) and by Egner (1996, 2000a,b), devoted to optimal plastic design of rotationally symmetric elements Received May 15, 2000 ˙ W. Egner and M. Zyczkowski Institute of Mechanics and Machine Design, Cracow University of Technology (Politechnika Krakowska), ul. Warszawska 24, PL 31-155 Krak´ ow, Poland e-mail:
[email protected]
and to plane heads of tension members. General approach to perturbation analysis of optimization problems with some reference to optimization in plasticity is presented in a recent monograph by Bonnans and Shapiro (2000). The problem of combined plastic torsion with bending (but without shear) was formulated by Handelman (1944) and Hill (1948, 1950). They used the Prandtl stress function. The relevant equation in terms of displacements ˙ was derived by Piechnik (1961), Piechnik and Zyczkowski (1961). Piechnik applied the perturbation method (but without boundary perturbation and optimal design) to obtain effective formulae for a circular cross-section. Advantages of numerical integration of Piechnik’s equation were pointed out by Miller and Malvern (1967). A more ˙ detailed survey is given by Zyczkowski (1981). The present paper uses three displacements and mean stress as the dependent variables. In comparison to Piechnik’s paper two essential generalizations are introduced: the effect of shear resulting in the loss of longitudinal homogeneity, and the boundary perturbation leading to optimal design of the bar. Plastic optimization of beams under bending with shear was considered by Heyman (1959) and Rozvany (1973, 1976) for some relatively simple cost functions. Rozvany and Hill (1976) proved that even for grillages in which the beams are permitted to resist torsion, the optimal solution often leads to torsion-free beams. However, in some structures or structural elements torsion cannot ˙ be eliminated and must be taken into account (Zyczkowski 1997, 1998). The present paper is devoted to bars or beams with prevailing torsion.
2 General perturbations for a plastic circular bar under torsion 2.1 Assumptions, basic solution As the basic solution (zero-th approximation) we consider a prismatic bar of the length under pure torsion. Optimal shape of the cross-section is then circular (within the class of solid, simply connected sections) or annular
395 (within the class of doubly connected sections). Subsequent optimization of annular shapes leads to thin-walled sections, but then a constraint imposed on the loss of stability of the wall must additionally be introduced. This problem will not be considered in the present paper: we simply assume that the ratio of the internal radius b to the external radius a is prescribed and the wall stability is not lost. The limit twisting moment M equals 2 M t = πτ0 a3 − b3 , 3
(1)
where τ0 is the yield-point stress in pure shear. For a solid section we may substitute b = 0. In the following we assume the material to be perfectly plastic, isotropic, incompressible, obeying the Huber–Mises–Hencky (HMH) yield condition. The strains are assumed to be small. Formally, the Hencky–Ilyushin deformation theory is used, but it may also be considered as the Levy–Mises theory of plastic flow if strains are replaced by strain rates. One end of the bar is assumed to be clamped and the other free. The cylindrical coordinate system r, θ, z is introduced with the origin in the centroid of free cross-section (Fig. 1). The moments are shown as vectors with double arrows. In the basic state the bar is under twisting moment Mz = M t which causes full yielding of the bar (limit load carrying capacity). The following equations show the non-zero stresses, strains and displacements, denoted by an additional subscript “0 ”: γzθ0 = ϑr , ϑ C= , 2τ0
τzθ0 = τ0 ,
ϕ0 =
uθ0 = −ϑr ( − z) ,
ϑr = Cr , 2τ0
2.2 Governing equations in the general case In the general case, under additional perturbing loadings which cause mainly bending and shear we obtain multiaxial state of stress and strain. Next we assume change of the shape of the bar in order to satisfy the yield condition. So, we can write the plasticity condition in the following form: 2
2 2 2 6 τrθ = 2σ02 , + τrz + τzθ
2
where σ0 =
√ 3τ0 .
(3)
The equations of internal equilibrium and the straindisplacement relations in cylindrical coordinates have the form ∂σr 1 ∂τrθ ∂τrz σr − σθ + + + =0, ∂r r ∂θ ∂z r ∂τrθ 1 ∂σθ ∂τzθ τrθ + + +2 =0, ∂r r ∂θ ∂z r ∂τrz 1 ∂τzθ ∂σz τrz + + + =0, ∂r r ∂θ ∂z r
εr =
∂ur , ∂r
εθ =
1 ∂uθ ur + , r ∂θ r
γrθ =
∂uθ uθ 1 ∂ur − + , ∂r r r ∂θ
γzθ =
1 ∂uz ∂uθ + . r ∂θ ∂z
(2)
where ϑ denotes unit angle of twist (arbitrary in the limit state), and ϕ is the plastic modulus.
2
(σr − σz ) + (σr − σθ ) + (σθ − σz ) +
(4)
εz =
γrz =
∂uz , ∂z
∂ur ∂uz + , ∂z ∂r (5)
The physical equations can be written in the following form: εk = ϕ (σk − δk σm ) ,
1 σm = σjj , 3
(6)
where δk denotes Kronecker’s symbol and Einstein’s summation convention is introduced. Five of the equations (6) are independent. Together with the incompressibility condition εr + εθ + εz = 0 ,
(7)
we obtain a set of 16 equations with 16 unknowns. According to the boundary perturbation method the unknowns (including perturbed shape a = a(θ, z) and b = b(θ, z) are expanded into series of certain small parameters αi , i = 1, 2, . . . , n, X= Fig. 1 Scheme of the bar
∞ ∞ j=0 k=0
...
∞ m=0
Xjk...m αj1 αk2 . . . αm n ,
(8)
396 where
2.3 Basic equations of the first correction
X = σr , σθ , σz , τrθ , τrz , τzθ , εr , εθ , εz , γrθ , γrz , γzθ , ur , uθ , uz , a, b
T
,
(9)
and individual parameters αi correspond to various perturbing loadings, for example to concentrated moment Mx , concentrated force P1 = P , uniformly distributed loading q, etc. In what follows, we confine our considerations just to one perturbing loading, namely concentrated force. The only small parameter, proportional to this force, will be simply denoted by α. Optimal design with perturbation due to concentrated bending moment (much simpler because of absence of shear and uniformity along the axis z) was considered by Bochenek et al. (1983). Moreover, the exact solution to this plastic op˙ timization problem, was derived by Zyczkowski (1997), who proved that the solution obtained earlier by Ober˙ weis and Zyczkowski (1997) for a certain family of elliptic sections satisfies the optimality condition. On the other hand, perturbation due to uniformly distributed loading can go along the lines of the present paper, being slightly more complicated. Some equations given below will be general, valid for any perturbing loading. Making use of the expansions (8) we realise that ten linear equations (4), (5), and (7) retain their form, just individual quantities are labelled with relevant additional subscripts. On the other hand, six remaining non-linear equations (3) and (6) are subject to changes resulting in their linearization. From (3) with substituted (2) we obtain τzθi = fi σk(i−1) , σk(i−2) , . . . , i = 1, 2, . . . , (10)
σr1 = σm1 +
ε r1 1 ∂ur1 = σm1 + , Cr Cr ∂r
σθ1 = σm1 +
1 u r1 εθ 1 1 ∂uθ1 = σm1 + + , Cr Cr2 ∂θ Cr r
σz1 = σm1 +
εz1 1 ∂uz1 = σm1 + , Cr Cr ∂z
τrz1 =
τrθ1
1 1 γrz = 2Cr 1 2Cr
1 1 = γrθ = 2Cr 1 2Cr
f1 = 0 , 1 2 2 (σr1 − σθ1 ) + (σθ1 − σz1 ) + 12τ0 2 2 2 (σz1 − σr1 ) + 6τrθ + 6τ , rz 1 1
f3 = −
1 2Cr
6τrθ1 τrθ2 + 6τrz1 τrz2 ] ,
(11)
and so on, and from (6) i j=0
ϕj σk(i−j) − δk σm(i−j) .
(12)
,
∂uθ1 uθ1 1 ∂ur1 − + ∂r r r ∂θ
∂uθ1 uθ1 1 ∂ur1 − + ∂r r r ∂θ
∂ur1 ∂uz1 + ∂z ∂r
.
(13)
+ ∂uθ1 + u r1 ∂θ
∂uθ1 uθ1 1 ∂ur1 − + ∂r r r ∂θ
∂uθ1 + u r1 ∂θ
∂σm1 1 ∂ + ∂z 2Cr ∂r
1 − Cr3
∂uθ1 uθ1 1 ∂ur1 − + ∂r r r ∂θ
1 ∂ Cr2 ∂θ
(σθ1 − σz1 ) (σθ2 − σz2 ) + (σz1 − σr1 ) (σz2 − σr2 ) +
∂σm1 1 ∂ 2 u r1 + + ∂r Cr ∂r2
∂σm1 1 ∂ + ∂θ 2C ∂r
1 [(σr1 − σθ1 ) (σr2 − σθ2 ) + 6τ0
∂ur1 ∂uz1 + ∂z ∂r
∂ur1 1 ∂uθ1 ur1 ∂uz1 + + + =0, ∂r r ∂θ r ∂z
1 ∂ 2Cr ∂z
f2 = −
Substituting the above relations into internal equilibrium equations (4) and incompressibility condition (7) we obtain the following set of 4 partial differential equations with 4 unknowns:
1 ∂ 2Cr2 ∂θ
where
εki =
On the first level of correction we obtain from (10) and (11) τzθ1 = 0. The equations of internal equilibrium are therefore simplified. Making use of (2), (5), and (12) we express the stresses in terms of displacements and mean stress σm1 as follows:
=0,
+
+
=0,
∂ur1 ∂uz1 + ∂z ∂r
+
1 ∂ 2 u z1 =0. Cr ∂z 2
(14)
The solution of the above set of equations depends on the additional loading of the bar (which is already preloaded by twisting moment). If we assume that loading acts in the yz plane, we can look for the solution in the following form:
397
u r1 =
∞
3 ∂ 4 frn ∂ 4 fzn ∂ frn ∂ 3 fzn 3 r + +r 3 2 +2 − ∂r3 ∂z ∂r2 ∂z 2 ∂r ∂z ∂r∂z 2
2 ∂ frn ∂ 2 fzn ∂ 2 fzn 2 2 + +4 =0. (19) r n 2 ∂r∂z ∂z 2 ∂r2 4
frn (r, z) sin (nθ) ,
n=1
uθ1 =
∞
fθn (r, z) cos (nθ) ,
n=1
u z1 =
∞
fzn (r, z) sin (nθ) ,
2.4 Boundary conditions
n=1
σm1 =
∞
fmn (r, z) sin (nθ) .
(15)
n=1
In the case of the horizontal or skew loadings the sine as well as cosine functions had to be introduced to the above equations, but for the concentrated force P1 shown in Fig. 1, (15) are sufficient. Substituting (15) into (14) we obtain ∂fzn ∂frn r + frn − nfθn + r =0, ∂r ∂z 2Cr
3 ∂fmn
∂r
+ 2r
2∂
2
frn ∂fθn − nr + 3nfθn − ∂r2 ∂r
2nCr2 fmn + r2
2Cr3
2
∂frn ∂ 2 fθn − 2n2 fθn + 2nfrn + nr =0, 2 ∂r ∂r
∂fmn ∂ 2 frn ∂ 2 fzn ∂ 2 fzn + r2 + r2 + 2r2 =0. 2 ∂z ∂r∂z ∂r ∂z 2
(16)
These equations may easily be reduced to a set of 2n equations by eliminating fθn , and fmn . Namely, from the first equation (16) we can calculate fθn fθn =
r ∂frn frn r ∂fzn + + , n ∂r n n ∂z
(17)
then from the third equation fmn = −
3 ∂ 2 frn r ∂ 3 frn 1 ∂ 2 fzn − − − 2 2 2 3 2Cn ∂r 2Cn ∂r Cn2 ∂z∂r
r ∂ 3 fzn 1 ∂frn 1 ∂fzn + + . 2Cn2 ∂z∂r2 2Cr ∂r Cr ∂z
(18)
As the result we obtain the following set of 2n equations with unknowns frn and fzn :
4
3 ∂ frn ∂ 4 fzn ∂ frn ∂ 3 fzn 3 r4 + + r 4 + 3 − ∂r4 ∂r3 ∂z ∂r3 ∂r2 ∂z
2 ∂ frn ∂ 2 frn ∂ 2fzn r n 2 + +2 + ∂r2 ∂z 2 ∂z∂r 2 2
n2 (n2 − 1)frn = 0 ,
aσr −
∂a ∂a τrθ − a τzr = 0 , ∂θ ∂z
aτrθ −
∂a ∂a σθ − a τθz = 0 , ∂θ ∂z
aτzr −
∂a ∂a τθz − a σz = 0 . ∂θ ∂z
(20)
Similar conditions hold for the internal surface r = b(θ, z) of a hollow bar. General expansion of (20) into power series of α allowing also for the expansion of a, was also given in the above-mentioned paper.
2 ∂ frn ∂ fzn n + 2 frn + r2 + r2 =0, ∂z 2 ∂r∂z 2
Boundary conditions for a contour in cylindrical coordinates r = a(θ, z) were derived by Egner et al. (1994). For a lateral surface free of loadings they may be written in the form
3 Effective solution for a solid bar under twisting moment and concentrated vertical force at the free end 3.1 The solution of governing equations for the first correction The differential equations (19) are close to ordinary differential equations of Euler’s type. Hence, we can look for the solution in the class of homogenous polynomials of mth-degree, where m can be any real or complex number. Here we restrict m to integers. Then the solution takes the form frn =
m ∞
(n)
Bi,m−i ri z m−i ,
m=0 i=0
fzn =
∞ m
(n)
Ci,m−i ri z m−i .
(21)
m=0 i=0
For a concentrated force it is sufficient to consider n = 1, 3; m = 0, 1, 2, 3, 4. For example, for n = 1 and m = 0, 1, we look for the solution in the form (1)
(1)
(1)
(1)
(1)
(1)
fr1 = B00 + B10 r + B01 z , fz1 = C00 + C10 r + C01 z ,
(22)
398
and obtain the following general solutions: (1) (1) (1) ur1 = B00 + B10 r + B01 z sin θ ,
(1) (1) (1) uz1 = C00 + C10 r + C01 z sin θ , (1)
(1) (1) εθ1 = − B10 + C01 sin θ ,
(3)
3 1 (3) z 2 (3) z B30 + 4C04 sin 3θ . 10 r r
(1)
(1)
γrθ1 = B10 cos θ , γzθ1 = σr1 =
(1) C00
1 Cr
r
(1)
(1)
γrz1 = B01 + C10 (1)
(1) z
+ C10 + C01
r
(1)
The requirement to satisfy the first of the above conditions for any θ and z gives eight relations:
cos θ , (3)
C04 = 0 ,
3 (1) (1) B + C10 sin θ , 2 10
(1)
τrθ1 =
1 (1) B cos θ . 2Cr 10
(1)
C02 = (3)
(23)
3.2 Boundary conditions for the first correction For the first correction the boundary conditions obtained from the expansions of (20) into power series take the form: σr1 |r=a0 = 0 , τrθ1 |r=a0 − τ0
(24)
In order to satisfy the first equation of (24) we write the formulae for σr1 using (23) and similar solutions for n = 1, m = 2, 3, 4 and n = 3, m = 0, 1, 2, 3, 4,
1 1 (1) (1) (1) (1) 2 σr1 = − 24B40 + 5C31 r + B + 8C03 r + C 2 30
(1) (1) 7 (1) 3 B10 B (1) 3 3C40 − B13 rz + + 01 + 2 2 r r
(3)
200 (3) C a0 , 3 13
C02 = −
1 (1) (1) C40 = − B13 , 4
20 (3) C , 9 12
51 (1) 2 3 (1) B a − B , 8 13 0 4 11
C01 = −
Similar solutions can be derived for higher coefficients n and exponents m. They will be given in Sect. 3.2.
∂a1 =0, ∂z ∂a1 a0 τrz1 |r=a0 − τ0 =0. ∂θ
(1)
(3)
B30 = −
C03 = B30 ,
(1) B10
1 sin θ , 2C r
1 (1) 1 (1) B + 2C01 sin θ , σz1 = Cr 2 10 1 (1) (1) τrz1 = B01 + C10 sin θ , 2Cr σθ1 = −
(25)
sin θ ,
+ B01
2 z (1) (1) z + 3 B30 − C03 + r r
40 (3) 20 (3) C (3) z C z + C13 z 2 + 01 + 2C02 + 9 12 3 r r
εz1 = C01 sin θ ,
1 (1) z 3 (1) C40 + B13 sin θ + 4 r 40 (3) 1 40 (3) 2 29 (3) (3) C13 r + B30 r + 348C04 rz + B20 + C 3 10 9
(1) (1) (1) (1) uθ1 = B00 + 2B10 r + B01 z + C01 r cos θ ,
εr1 = B10 sin θ ,
3 (1) (1) B + 2C02 2 11
40 (3) (3) B a0 + 180C13 a30 , 9 20
3 (1) 17 (1) (1) (1) (1) B01 = − B10 − B30 a20 + 24B40 a30 + 5C31 a30 . 2 2
(26)
In the remaining two conditions we have the same function a1 . In order to satisfy both of them the following condition (Schwarz condition) must hold: ∂τrz1 ∂τrθ1 = a0 . (27) ∂θ r=a0 ∂z r=a0 The formulae for τrθ1 and τrz1 are written below:
3 (1) 2 1 (1) (1) (1) τrθ1 8B40 + C31 r + B − 4C03 r + 2 2 30
9 (1) 1 (1) (1) (1) B − 3C40 rz + B20 − B02 + 2 13 2 1 = C
(1) 3 (1) 87 (1) 1 B10 (1) (1) B21 − B03 z + C31 + 120B40 z 2 + + 2 4 2 r 2 1 3 1 (1) z (1) (1) z (1) z B11 + B30 − 2C03 + B13 cos θ + 2 r r 2 r 1 27 (3) 7 (3) (3) (3) 20C13 r2 + B30 r + 324C04 rz + B20 + C 10 3 7 (3) 7 (3) C12 z + C13 z 2 cos 3θ , 3 2
(28)
399
1 (1) 2 (1) τrz1 3C40 − B13 r + 2
21 (1) (1) C + 48B40 rz + 2 31
3 (1) 1 (1) (1) (1) 2B21 + B03 r − 4C02 + B11 + 2 2
2π
(1) (1) (1) (1) 2 B30 + 4C03 z + 6 B13 − C40 z 2 +
3¯ a1 a ¯23 cos2 θ cos2 3θ + a ¯33 cos θ cos3 3θ dθ = 0 .
1 (1) 1 (1) z (1) 1 (1) B 01 + B10 + B − B20 + 2 r 2 02 r
2
3 1 3 (1) 29 (1) z (1) (1) z B03 − B21 − C31 + 40B40 sin θ + 2 2 r 4 r 1 20 (3) 20 (3) (3) (3) 20C04 r2 − C12 r − C13 rz − 4C02 − C 9 3
From the above equation we obtain a11 (z) = 0, and a13 (z) may be an arbitrary constant. So we obtain the additional conditions:
1 = C
(3) C10
2 (3) 1 (3) B z − 24C04 z 2 + 5 30 2 r
Substituting (31) into (32) we obtain the following integral: 2 3¯ a0 a ¯1 cos2 θ + a ¯31 cos4 θ + 6¯ a0a ¯1 a ¯3 cos2 θ cos 3θ +
0
¯3 cos3 θ cos 3θ + 3¯ a0a ¯23 cos θ cos2 3θ + 3¯ a31 a
+
(1)
B40 = −
29 (1) C , 160 31
3 (1) (1) B21 = B03 , 2 1 (1) 96 (1) (1) B20 = B02 + C31 a20 , 2 5 (1)
1 (3) z 2 1 (3) z (3) z B20 + C12 + C13 r 2 r 2 r
(1)
(1)
B01 = −B10 − 9B03 a20 ,
3 sin 3θ .
(29)
Fulfilment of (27) with (26) taken into account leads to the following relations: (3)
(3)
(3)
(3)
(1)
(1)
(3)
(3)
(1)
(1)
(1)
(33)
(34)
and the function describing the contour in the first approximation has the form (3)
a = a0 − α
C10 cos 3θ . 6Cτ0
(35)
(1)
C13 = B30 = C12 = C02 = B30 = C03 = B13 = 3.4 The integral conditions of internal equilibrium
C40 = B20 = C01 = B11 = C02 = 0 , (1)
(1)
(1)
B10 = −112B40 a30 − 24C31 a30 .
(30)
Then the first correction of the contour equals a1 (θ, z) = a11 (z) cos θ + a13 (z) cos 3θ ,
(31)
where a11 (z) and a13 (z) are polynomials of the third degree.
In this section we require the sectional forces resulting from the stress distribution to be equal to those resulting from external loadings. In the case under consideration we have three kinds of forces (the condition on normal force is satisfied identically): bending moment, twisting moment and shear force. These conditions can be written in the following form: 2πa ∞
3.3 The condition of invariant location of the centroid Now we demand from the first correction not to cause change of location of centroid, otherwise the additional eccentric twisting would appear. The radius vector describing the contour of the bar in first correction is given by (31) with added a0 . Then the static moment with respect to the vertical axis y which is required to vanish we can express in the form 2 Sy = x dA = [a(θ)] cos θ dr dθ = 1 3
2π 3
[a(θ)] cos θ dθ . 0
(32)
0
αk (τrzk sin θ + τzθk cos θ) r dr dθ = αt ,
(37)
αk τzθk r2 drdθ = Mz = M t ,
(38)
0 k=0
2πa ∞ 0
(36)
0 k=0
2πa ∞ 0
αk σzk r2 sin θ dr dθ = αtz ,
0 k=0
where αt = P and a = a(z, θ). M t denotes load-carrying capacity for twisting of circular cross-section. The upper limit of integration a is subject to expanding into series: ∞ a(z, θ) = i=0 ai αi . In the first condition of (36) appears the stress σz . Taking into account the relations (26), (30), and (34) we obtain the following formula for σz1 :
400
384 (1) 2 43 (1) 2 (1) (1) C31 r − 3B02 − C a − 9B03 z − 40 5 31 0 9 (1) 2 211 (1) 3 1 C31 z + C31 a0 sin θ . (39) 5 20 r
σz1
1 = C
Substituting the above formula into the condition (36) we obtain the following relations: (1)
(1)
C31 = B02 = 0 ,
(1)
B03 = −
Ct . 3πa30
(40)
Taking them into account we obtain from (29) τrz1 =
3t 2πa30
(3) a20 1 C10 − r sin θ + sin 3θ . r 2 r
(41)
Substituting the above result into equations (37) and (38) we find them satisfied identically. Since τzθ1 = 0, there is no additional twisting for the first correction.
and leaving the term with α2 we find that the minimum (3) is obtained for C10 = 0 and in the following we assume a1 = 0. 3.6 List of formulae for the first correction for solid bar Finally for the first correction we have the following formulae: σr1 = σθ1 = 0 ,
σz1 =
τrθ1 = τzθ1 = 0 ,
τrz1
a1 = 0 ,
ϕ1 =
3t z sin θ , πa30
2 3 t a0 − r sin θ , = 2 πa30 r
Ct 2 a0 − r2 cos θ . 3 6πa0 τ0
(45)
3.7 The second correction for a solid bar
3.5 The Drucker–Shield condition
For the second correction the yield condition (3) gives At the end we demand to satisfy the necessary optimality condition of the contour shape. This condition is known in literature as the Drucker–Shield condition. According to it, if the boundary of a structure in the plastic state is not loaded, then the necessary condition of optimality of the shape is that the unit power dissipated in the Levy– Mises theory or energy dissipated in the Hencky–Ilyushin theory on that boundary is constant. It can be written in the following form: D = σij εij |S = const .
(42)
For the first correction in view of τzθ1 = 0, it has the following shape: D = τ0 γzθ0 |r=a0 +
∂γzθ0 α τ0 γzθ1 |r=a0 + τ0 a1 + . . . = const . ∂r r=a0
(43)
The above condition results in the following relations: (1)
(1)
C00 = 8B03 a30 ,
8 (3) (3) C00 = − C10 a0 . 9
(44)
Finally, we must require to satisfy the boundary condition connected with clamping of the bar: uθ (r = a0 ; θ = 0, π; z = ) = 0, uz (r = a0 ; θ = ±π/2; z = ) = 0. These conditions give certain relations involving the constants (1) (1) B10 , and B00 not yet used. The conditions are not shown here, because these constants will not appear in the fur(3) ther analysis. On the other hand, the constant C10 in (35) is not determined by the Drucker–Shield condition in the linear approximation (43). However, calculating the cross-sectional area by integration of (35) squared
2 σz21 + 3τrz + 6τ0 τzθ2 = 0 . 1
Making use of (45) we present τzθ2 in the form
2 3 t2 3 a20 2 τzθ2 = − r + z (cos 2θ − 1) . 4 π 2 a60 τ0 4 r
(46)
(47)
Similarly as for the first correction the number of equations can be reduced by elimination of stresses and strains by displacements and mean stress. Making use of (4)–(7) and (45) we eliminate the remaining stress components
1 H 3a20 r 1 ∂ur2 σr2 = σm2 + − z sin 2θ + , 4 τ0 r 3 Cr ∂r
1 H 3a20 r 1 ∂uθ2 σθ2 = σm2 + − z sin 2θ + + 4 τ0 r 3 Cr2 ∂θ 1 ur , Cr2 2
1 H 3a20 r 1 ∂uz2 σz2 = σm2 − − z sin 2θ + , (48) 2 τ0 r 3 Cr ∂z ∂ur2 ∂uz2 + − ∂z ∂r
3 H r2 2 2 a40 − a0 + 2 sin 2θ , 8 τ0 3 3 3r
∂uθ2 uθ2 1 ∂ur2 1 τrθ2 = − + . 2Cr ∂r r r ∂θ τrz2 =
1 2Cr
(49)
In the above equations H is proportional to the concentrated force squared, H = t2 /π 2 a60 . Substituting these stresses to the equilibrium equations, together with incompressibility condition we ob-
401 tain the following set of four linear nonhomogenous partial differential equations ∂ur2 1 ∂uθ2 ur2 ∂uz2 + + + =0, ∂r r ∂θ r ∂z ∂σm2 1 ∂ 2 uθ2 1 ∂ 2 u r2 3 ∂uθ2 + + − + 2 ∂r Cr ∂r 2Cr2 ∂r∂θ 2Cr3 ∂θ ur 1 ∂ 2 u r2 1 ∂ 2 u z2 1 ∂ 2 u r2 + + − 23 = 3 2 2 2Cr ∂θ 2Cr ∂z 2Cr ∂r∂z Cr
1 H 1 3a20 + 2 z sin 2θ , 4 τ0 3 r 1 ∂σm2 1 ∂ 2 uθ2 1 ∂ur2 1 ∂ 2 uθ2 + + + + r ∂θ 2Cr ∂r2 Cr3 ∂θ2 Cr3 ∂θ
3H 8 a20 1 ∂ 2 u r2 = z 1− + cos 2θ , 2Cr2 ∂θ∂r 2 τ0 9 r2
∂σm2 1 ∂ ∂ur2 ∂uz2 1 ∂ 2 u z2 + + + = ∂z 2Cr ∂r ∂z ∂r Cr ∂z 2
H 4 a2 3 z 2 a40 r− 0 + + 3 sin 2θ . τ0 3 r 2 r r
CH 3 r z, τ0
∂ 4 f r2 ∂ 4 f z2 ∂ 3 f r2 ∂ 3 f z2 4 3 r + +r 3 2 +2 − ∂r3 ∂z ∂r2 ∂z 2 ∂r ∂z ∂r∂z 2
∂ 2 f r2 ∂ 2 f z2 ∂ 2 f z2 2 r 8 + 16 +4 = ∂r∂z ∂z 2 ∂r2
CH 1 2 2 3 8 a0 r − 2r4 − r2 z 2 − a40 . (53) τ0 4 2 −6
We assume the following form of particular solution for the functionf r2 : (50)
The particular solution can be searched within the functions: ur2 = f r2 (r, z) sin 2θ , uθ2 = f θ2 (r, z) cos 2θ + uz2 = f z2 (r, z) sin 2θ ,
1 CH 3 r z, 2 τ0 σm2 = f m2 (r, z) sin 2θ .
(51)
Hence we obtain the following equations:
8 a20 + 9 r2
(54)
Now, the function f z2 must satisfy simultaneously two equations:
∂ 3 f z2 ∂ 2 f z2 ∂f z2 ∂ 4 3 2 r + 3r − 8r =0, ∂z ∂r3 ∂r2 ∂r
∂ 2 f z2 ∂f z2 ∂ 2 f z2 ∂2 4 3 2 + 2r = r − 16r f z2 − 4r2 2 2 ∂z ∂r ∂r ∂r2
CH 1 2 2 3 2 2 4 4 8 a r − 2r − r z − a0 . (55) τ0 4 0 2
1 CH 1 CH , D04 = , 3 τ0 16 τ0 CH 4 1 1 CH 2 D = −2 a0 , D02 = − D20 − a . τ0 4 16 τ0 0 D40 =
,
1 ∂ 2 f r2 1 ∂ 2 f z2 ∂f m2 1 ∂ 2 f z2 + = + + 2Cr ∂z∂r 2Cr ∂r2 ∂z Cr ∂z 2
H 4 a2 3 z 2 a40 r− 0 + + 3 , τ0 3 r 2 r r ∂f r2 2 fr2 ∂f z2 − f θ2 + + =0. ∂r r r ∂z
1 CH 3 r z. 2 τ0
then the first equation (55) is satisfied automatically, and the second one results in
2 1 ∂ 2 f θ2 2 f m2 + + f r2 − 2f θ2 + 2 3 r 2Cr ∂r Cr
f r2 =
If the particular solution will be written in the following shape: r f z2 = D40 r4 + D20 r2 + D02 z 2 + D04 z 4 + D ln , (56) a0
∂f m2 1 ∂ 2 f r2 1 ∂f θ2 3 + − + f − f r2 + ∂r Cr ∂r2 Cr2 ∂r Cr3 θ2
∂f r2 ∂ 2 f z2 1 1H 1 3a20 + = z + 2 , 2Cr ∂z 2 ∂r∂z 4 τ0 3 r
1 ∂f r2 3H =− z Cr2 ∂r 2 τ0
Similarly as for the first correction we eliminate f θ2 and f m2 and obtain the following two equations:
4 4 3 3 ∂ f ∂ f ∂ f ∂ f r z r z 2 2 + 3 2 + r3 4 +3 2 2 − r4 ∂r4 ∂r ∂z ∂r3 ∂r ∂z
∂ 2 f r2 ∂ 2 f r2 ∂ 2 f z2 4r2 8 + 4 + 8 + 12f r2 = ∂r2 ∂z 2 ∂z∂r
(52)
(57)
Finally we have the particular solution in the following form: 1 f r2 = H1 r3 z , 2
1 1 1 2 2 f z2 = D20 r − D20 + H1 a0 z 2 + H1 r4 + 4 4 3 1 r H1 z 4 − 2H1 a40 ln . (58) 16 a0
402 where the new notation has been introduced: H1 = CH/τ0 = Ct2 /τ0 π 2 a60 . Now, we are looking for the general solution of the homogeneous part of (53). We restrict general solution to the following functions: ur2 = fr2 (r, z) sin (2θ) ,
uθ2 = fθ2 (r, z) cos (2θ) ,
uz2 = fz2 (r, z) sin (2θ) ,
σm2 = fm2 (r, z) sin (2θ) . (59)
Using the way shown in (21) we assume f r2 =
4 4
Ei(m−i) ri z m−i ,
m=0 i=0
fz2 =
4 4
Fi(m−i) ri z m−i ,
(60)
m=1 i=1
and obtain
6 3 1 E20 r2 + D20 r2 − E30 r2 z − E31 r2 z 2 − E20 rz − 5 5 2
1 1 1 1 2 3 2 E21 rz − E22 rz + F01 z − D20 + H1 a0 z 2 − 4 6 4 4 1 1 1 1 E20 z 2 + E30 z 3 + H1 z 4 + E31 z 4 − 4 10 16 40 r 2H1 a40 ln sin 2θ , a0
1 79 9 9 17 σm2 = − E22 r2 − E30 r − E31 + H1 rz − C 72 5 5 12
1 1 1 1 E20 − E21 z − E22 z 2 − E20 + D20 + 8 8 8 2 7 3 z2 2 z H1 a0 + E30 + 4 r 10 r
3 1 1 1 z E31 + H1 sin 2θ . (62) 2 5 2 r
2 fr2 = E22 r4 + E30 r3 + E20 r2 + E21 r2 z + E22 r2 z 2 + 9 E31 r3 z + E00 ,
3.8 Boundary conditions for the second correction
1 1 1 fz2 = E31 r4 − E21 r3 − E22 r3 z + F20 r2 − 4 8 4
Boundary conditions for the second correction take the following form resulting from expansion of (20):
6 3 1 1 E30 r2 z − E31 r2 z 2 + F10 r − E20 rz − E21 rz 2 − 5 5 2 4
σr2 |r=a0 = 0 ,
1 1 1 1 E22 rz 3 + F01 z − F20 z 2 + E30 z 3 + E31 z 4 . 6 4 10 40
τrθ2 |r=a0 − τ0
(61)
Finally, the second corrections of displacements and mean stress have the form: 2 1 u r2 = E22 r4 + E30 r3 + E31 r3 z + H1 r3 z + 9 2
E20 r2 + E21 r2 z + E22 r2 z 2 + E00 sin 2θ , uθ2 =
31 7 7 E22 r4 + E30 r3 + E31 + H1 r3 z + 72 5 5
5 2 r (E20 + E21 z + E22 z 2 ) − 4
1 1 3 E20 + D20 + H1 a20 rz + E30 rz 2 + 4 4 20
1 1 1 1 1 E31 + H1 rz 3 + E00 cos 2θ + H1 r3 z , 4 5 2 2 2 1 1 1 1 uz2 = E31 r4 + H1 r4 − E21 r3 − E22 r3 z + 4 3 8 4
∂a2 =0, ∂z
∂τrz1 a0 a1 + a0 τrz2 |r=a0 + a1 τrz1 |r=a0 − ∂r r=a0 τ0
∂a2 =0. ∂θ
(63)
The first equation of the set (63) is of the same type as for the first correction. In the third equation the terms connected with a1 are equal to zero and from the Schwarz condition for a2 we obtain similar equation as for the first correction: ∂τrθ2 ∂τrz2 = a . (64) 0 ∂θ r=a0 ∂z r=a0 Since the way of satisfying the boundary conditions for the second corrections is the same as for the first one, we will not give here the details and write the final form of the second correction for the external contour:
H1 a0 17 2 F10 5 2 a2 = a − − z cos 2θ + Cτ0 48 0 4H1 a0 8 H1 a0 z 2 + Ca . Cτ0 4
(65)
403 The integral equations of the internal equilibrium for shear forces and bending moments are satisfied automatically. The equilibrium condition for the twisting moment which can be written in the following form:
2πa
Mz =
τzθ r dA = 0
A
2P 2 z 2 ax = 1− +... , a0 9Mt2
τ0 + α2 τzθ2 r2 dr dθ =
0
M t + 0 · α + 0 · α2 + . . .
(66)
(where a = a0 + α2 α2 ) allows us to calculate constant Ca : Ca =
3 H1 a30 . 10 C1 τ0
(67)
Now, the formula for a2 can be written in the following form:
H1 a0 F10 5 2 17 2 a2 = a − − z cos 2θ + Cτ0 48 0 4H1 a0 8 1 2 3 2 z + a0 . (68) 4 10 We should satisfy the Drucker-Shield condition. For the second correction it can be written in the form (non-zero terms only): (σz1 εz1 + τ0 γzθ2 + τzθ2 γzθ0 + τrz1 γrz1 )r=a0 + 2Cτ02 a2
= const .
(69)
It turns out that this condition can be satisfied only approximately (terms of type z 0 cos 2θ only). Using it we can calculate the constant F10 . F10 = −
13 H1 a30 . 12
(70)
After taking into account the above constant, defining the dimensionless small parameter in the form: α=α
t , πa20 τ0
used this shape as an approximation if the bending moment is variable along the axis z. As a result the ellipse with the axes given by the following relation:
(71)
and taking into account the results of the first correction we can write the function describing lateral surface in the following dimensionless form:
a z2 1 z2 3 2 5 = 1+α 1 − 2 cos 2θ + + . (72) a0 8 a0 4 a20 10 The conditions of support for the second correction are of the same form as for the first correction: uθ2 (r = a0 ; θ = 0, π; z = ) = 0, uz2 (r = a0 ; θ = ±π/2; z = ) = 0. The condition for displacement uz2 is satisfied automatically, thus the condition for ur2 makes it possible to calculate the constant E00 . At the end we compare the obtained shape with the exact optimal shape of a bar under torsion and bending ˙ in plastic range obtained by Zyczkowski (1998), who considered the similar problem but without shear effect and
ay 4P 2 z 2 = 1+ +... , a0 9Mt2
(73)
was obtained. Axes ax and ay are placed in the plane z=const (Fig. 2), and a0 is a radius of a bar under pure torsion at the stage of collapse: a0 =
3
3Mt . 2πτ0
(74)
When using boundary perturbation method with shear taken into account, the axes of a cross-section are given by the following formulae resulting from (72): 37 P 2 a20 3 P 2z2 ax = 1+ − +... , a0 90 Mt2 18 Mt2 13 P 2 a20 7 P 2z2 ay = 1− + +... . a0 90 Mt2 18 Mt2
(75)
Comparing the relations (73) and (75) we can see that in equation (73) all correcting terms depend on z. It is the result of neglecting the effects of shear, taken into account in (75). Bending effect, which is described by the terms proportional to z 2 is for both solutions similar.
3.9 Numerical example The example was calculated for /a0 = 4 and α ¯ =0.15. The following interpretation for the above data can be presented. The value of small parameter α ¯ can be understood as a ratio of bending moment in clamped cross-section Mb to the torque: Mb Mt
=
αt 2 3 3 πτ0 a0
=
3 α ¯. 2 a0
(76)
For the above data this ratio is equal Mb /M t = 0.9. Figure 2 shows the corresponding optimal shape. Figure 3 shows the shape of the longitudinal contour of the bar, obtained as sections in the planes xz and yz. It is seen from the third of equations (11) that the third correction governed by α3 must contain the terms with cos 3θ which cannot be avoided as in the first correction. The symmetry of the cross-section with respect to the axis y is then lost. However, the calculations of this correction are very cumbersome and will not be given here.
404
Fig. 2 Optimal shape of the solid bar
Fig. 3 Variation of semiaxes along the bar
4 Effective solution for a hollow bar 4.1 Boundary conditions for the first correction for a bar with annular cross-section For bending and torsion the layers which are close to the axis of symmetry of a bar are not appropriately exploited. So we may expect further profits for annular cross-section of bars (doubly connected cross-sections). Such an approach was applied by Bochenek et al. (1983). In the case of an annular cross-section of a bar the form of the boundary conditions given by (20) and their expansion into power series do not change. However, we must remember that they must be satisfied at both boundaries a(z, θ) and b(z, θ). It turns out that the exact fulfilment of all equations is impossible. Therefore we demand for these equations to be satisfied identically for the terms z 0 and z 1 . For the terms with z 2 and z 3 which ap-
pear for example in the equation for σr1 , the fulfilment is possible if we take into account higher powers of variables r and z. On the other hand, taking into account the higher powers makes impossible to satisfy the boundary conditions for the terms connected with these higher powers. It is clear that we can find the solution with any precision we want, but always the terms with highest powers would not fulfil some equations. Hence, the obtained solution can be understood as an approximate solution. This approximation is better if we consider sections closer to the free end of a bar. For z → 0 the solution leads to the exact one. It should be mentioned that in the case of the annular cross-section under consideration, only the solution connected with trigonometric functions of a single angle θ will be taken into account. The calculations are more complicated here, since we have two independent surfaces to design. This justifies neglecting the terms with triple angle. The experiences from the previous problem show that these terms make only the volume bigger. Moreover the analysis for a single angle gives chances to find a nontrivial correction for a = a(r, z) and b = b(r, z) as early as for the first correction. We have two independent surfaces to design, so the rigid movement of these surfaces described by the solution with trigonometric functions of a single angle is possible. The condition of invariant location of centroid will not cause a zero solution. Then further considerations differ from the previous one only in the range of neglected terms with triple angle, fulfilling of boundary conditions in the terms with z 0 and z 1 and the necessity of satisfying the boundary conditions at both lateral surfaces. After satisfying boundary conditions at both surfaces we obtain the following relations for the first correction of the shape:
1 3 a1 = −3B13 a30 − 2B21 + B03 a20 − cτ0 2 5 1 B11 a0 − (B01 + B10 ) cos θ , 2 2
1 3 3 −3B13 b0 − 2B21 + B03 b20 − b1 = cτ0 2 5 1 B11 b0 − (B01 + B10 ) cos θ . (77) 2 2 The next step is demanding the invariant location of centroid. This condition can be written in the following form: 1 Sy = 3
2π
3 a − b3 cos θ dθ =
0
1 3
2π
3 a0 + 3a20a1 +
0
. . . − b30 − 3b20b1 − . . . cos θ dθ = 0 ,
(78)
405 where
4.2 Numerical example
a1 = Ca cos θ , b1 = Cb cos θ .
(79)
The fulfilment of the above condition results in the following relation: a20 Ca − b20 Cb = 0 .
(80)
Next, we demand of satisfying the condition of the bending moment equilibrium. In the first approximation it has the form shown below:
Figure 4 shows the shape of the contour for the small parameter α ¯ = 0.15 and ratio of radii k=2. It is seen, that as a result of design the wall at the right side is thicker, and at the left side is thinner. It is caused by the accumulation of shear stresses at the right-hand side (from torsion and shear). Figure 5 shows the relation between dimensionless area (85) and the ratio of radii k.
2πa0 σz1 r2 sin θ dr dθ = tz .
(81)
0 b0
The remaining conditions, describing the equilibrium of the twisting moment and shear force, are satisfied identically. The Drucker-Shield conditions are the same as for the previous example, but they have to be satisfied at both surfaces
∂γzθ0 γzθ1 + a1 = const , ∂r r=a0
∂γzθ0 γzθ1 + b1 = const . (82) ∂r r=b0 After the above considerations we obtain the solution with two arbitrary constants. However, detailed analysis has shown that both of the constants in linear way increased the dimensions of the bar. Therefore in further considerations we assume them equal to zero. Finally we can write the first corrections of the shape in the following form: a1 =
t 3 b20 cos θ , 2 b30 − a30 τ0 π
b1 =
t 3 a20 cos θ . 2 b30 − a30 τ0 π
Fig. 4 Optimal cross-section of a hollow bar
(83)
We define the dimensionless small parameter in the same way as for the first example (71) introduce the following notation: a0 /b0 = k, and obtain the following formulae: a 3 k = 1+α cos θ , a0 2 (1 − k 3) b 1 3 k3 = +α cos θ . a0 k 2 (1 − k 3)
(84)
The following relation gives the dimensionless area of a cross-section: 1 − k4 A 29 = 1+α (85) . A0 8 1 − 12 1 − k 2 2 k
k
In the first approximation it is constant along the axis of the bar. The second correction will not be given here.
Fig. 5 Dimensionless cross-sectional area in terms of the ratio of radii k
406 Three curves are shown which correspond to various values of small parameter α ¯ , that means to various contributions of bending and shearing in relation to the basic load (twisting). The above picture shows that the greater is the perturbation (force P ) the greater are the profits in volume.
5 Conclusions 1. Effective solutions for optimization of a bar under torsion with superposed bending and shear are obtained via the boundary perturbation method. 2. The first correction of cross-sectional shape vanishes for a solid bar, but is different from zero for a hollow bar. 3. The more thin-walled is the doubly connected crosssection, the greater is the profit in volume. However, we should remember that for thin-walled sections the condition of stability should be introduced. Such a condition was not considered in the present paper.
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